# 2nd PUC Maths Previous Year Question Paper June 2019

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## Karnataka 2nd PUC Maths Previous Year Question Paper June 2019

Time: 3 Hrs 15 Min
Max. Marks: 100

Part – A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.
Let * be the binary operation on N given.by a * b = LCM of a and b, ∀ a ? b ∈ N.
Given a * b = LCM of a and b ∀ a, b ∈ N 5*7 = 35

Question 2.
Find the principal value of cot-1$$\left(-\frac{1}{\sqrt{3}}\right)$$
Cot-1$$\left(-\frac{1}{\sqrt{3}}\right)$$ = π – Cot-1$$\left(-\frac{1}{\sqrt{3}}\right)$$
[∵ cot-1(-x) = π – cot-1(x)]
= π – $$\frac{\pi}{3}$$
∴ cot-1 $$\left(\frac{-1}{\sqrt{3}}\right)=\frac{2 \pi}{3}$$

Question 3.
If A = [aij] where elements are given by aij = $$\frac { 1 }{ 2 }$$|-3i + j| construct 2 × 2 matrix.
Given aij = $$\frac { 1 }{ 2 }$$ |-3i + j|
a11 = $$\frac { 1 }{ 2 }$$|-3(1) + 1|= $$\frac { 1 }{ 2 }$$|-2| = 1
a12 = $$\frac { 1 }{ 2 }$$ |-3(1) + 2| = $$\frac { 1 }{ 2 }$$|-l| = $$\frac { 1 }{ 2 }$$
a21 = $$\frac { 1 }{ 2 }$$ |-3(2) + 2| = $$\frac { 1 }{ 2 }$$ |-5| = $$\frac { 5 }{ 2 }$$
a22 = $$\frac { 1 }{ 2 }$$ |-3(2) + 2| = $$\frac { 1 }{ 2 }$$ |-4| = 2

Question 4.
Find a value of x for which  3 – x2 = 3 – 8 ⇒ x2 = 8 ⇒ x = ±2$$\sqrt{2}$$ Question 5.
If y = cos-1 (ex), find $$\frac{d y}{d x}$$.
y = cos-1 (ex) (diff wrt x) Question 6.
Find $$\int \sec ^{2}(7-4 x) d x$$ Question 7.
If â = $$\frac{1}{\sqrt{14}}$$ (2î + 3ĵ + k̂), then write the directions cosines of â.
â = $$\frac{1}{\sqrt{14}}$$ (2î + 3ĵ + k̂)
∴ Directions cosines are :
$$\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}$$

Question 8.
Find the intercepts cut off by the plane 2x + y – z = 5
2x + y – z = 5 …..(1) ÷ 5
$$\frac{2 x}{5}+\frac{y}{5}-\frac{z}{5}=1$$
or $$\frac{x}{(5 / 2)}+\frac{y}{5}+\frac{z}{(-5)}=1$$
∴ x – intercept = $$\frac { 5 }{ 2 }$$ ; y – intercept = 5, z – intercept = -5

Question 9.
Define feasible region in a linear programming problem.
Feasible region is defined as the common region determined by all the constraints including non – negative constraints x ≥ 0, y ≥ 0 of a linear programming problem is called as Feasible region.

Question 10.
If P(A) = $$\frac { 3 }{ 5 }$$ and P(B) = $$\frac { 1 }{ 5 }$$ find P(A ∩ B), where A and B are independent events.
P(A) = $$\frac { 3 }{ 5 }$$; P(B) = $$\frac { 1 }{ 5 }$$
P(A ∩ B) = P(A).P(B)
[Since A and B are independent events]
= $$\frac{3}{5} \cdot \frac{1}{5}$$ = $$\frac { 3 }{ 25 }$$

Part-B

Answer any TEN questions: (10 x 2 = 20)

Question 11.
Find g.f and f.g, if (x) = 8x3 and g(x) = x1/3
Given
f(x) = 8x3 ,g(x)= x1/3
fog (x) = f(g(x))
= f(x1/3) =8(x3)1/3 =8x
gof (x) = g(f(x)) = g(8x3)
= (8x3)1/3 = 2x Question 12.
Prove that tan-1 $$\frac { 2 }{ 11 }$$ + tan-1 $$\frac { 7 }{ 24 }$$ = tan-1 $$\frac { 1 }{ 2 }$$
Consider LHS : tan-1 ($$\frac { 2 }{ 11 }$$) + tan-1 ($$\frac { 7 }{ 24 }$$) Question 13.
Write cot-1 $$\left(\frac{1}{\sqrt{x^{2}-1}}\right)$$ ,x > 1 in the simplest form.
Let x = sec q ⇒ q = sec-1 x.
Now cot-1 $$\left(\frac{1}{\sqrt{x^{2}-1}}\right)$$ = cot-1 $$\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)$$
= cot-1 $$\left(\frac{1}{\tan \theta}\right)$$ = cot-1(cot θ) = θ = sec-1 x.

Question 14.
Find the area of the triangle with vertices (3,8), (-4, 2) and (5, 1) using determinants.
Area of triangle = $$\frac { 1 }{ 2 }$$ [3(2 – 1) -8(-4 -5) + 1(-4 -10)]
= $$\frac { 1 }{ 2 }$$ [3 + 72 -14] = $$\frac { 61 }{ 2 }$$sq.units

Question 15.
Find $$\frac{d y}{d x}$$ if x2 + xy + y2 = 100.
x2 + xy + y2 = 100 (diff wrt x)
2x + x.$$\frac{d y}{d x}$$ + y.1 + 2y$$\frac{d y}{d x}$$ = 0
x.$$\frac{d y}{d x}$$ + 2y$$\frac{d y}{d x}$$ = 2x – y
$$\frac{d y}{d x}$$(x + 2y) = -(2x + y)
$$\therefore \frac{d y}{d x}=\frac{-(2 x+y)}{x+2 y}$$ Question 16.
Find $$\frac{d y}{d x}$$ if y = (logx)cosx.
Let y = (log x)cos x
⇒ log y = cos x log (log x)
⇒ $$\frac{d }{d x}$$logy = $$\frac{d }{d x}$$cosx log(logx) Question 17.
Find the interval in which the function f given f(x) = 2x2 – 3x is strictly increasing.
f(x) = 2x2 – 3x
diff wrt x
f1 (x) = 4x – 3
f1 (x) = 0 ⇒ 4x – 3 = 0
or 4x = 3
⇒ x = 3/4
∴ Point x = 3/4 divides the real line into two disjoint intervals (- ∞, 3/4) and (3/4, ∞) In the interval (3/4, ∞), f1(x) = 4x – 3 > 0
∴ f is strictly increasing in (3/4, ∞)

Question 18.
Find $$\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x$$   Question 19.
Integrate x sec2 x with respect to x. $$\int x \sec ^{2} x d x=x \int \sec ^{2} x d x-\int\left(\int \sec ^{2} x\right) \cdot \frac{d}{d x}(x) \cdot d x$$
= x. tan x – $$\int \tan x \cdot 1 \cdot d x$$
= x . tanx – log |sec x| + c

Question 20.
Find the order and degree (if defined) of the differential equation :
y111 – y2 + e3 = 0
Order = 3, Degree = not defined.

Question 21.
If $$\overrightarrow{\mathrm{a}}$$ is a unit vector and $$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})$$ = 8, then find $$|\overrightarrow{\mathrm{x}}|$$ . Question 22.
Find the area of the parallelogram whose adjacent sides are given by the vectors $$\overrightarrow{\mathrm{a}}$$ = 3î + ĵ + 4 k̂ and $$\overrightarrow{\mathrm{b}}$$ = î – ĵ + k̂
Area of parallelogram = $$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|$$  Question 23.
Find the angle between the pair of lines $$\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$$ and $$\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}$$
Given
$$\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$$
Here $$\overrightarrow{\mathrm{b}}_{1}$$ = 3î + 5ĵ + 4k̂
and $$\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}$$ ⇒ î + ĵ + 2k̂ Question 24.
The probability distribution of random variable X is as follows : Here E(X) = Σxi P(xi)
$$=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}=\frac{32+2}{221}=\frac{34}{221}$$

Part-C

Answer any TEN questions: (10 × 3 = 30)

Question 25.
Determine whether the relation R in the set A = {1, 2, 3, …., 13, 14} defined as R = {(x,y); 3x – y = 0} is reflexive, symmetric and transitive.
Given A = {1,2,3, ….,13, 14}
R = {(x, y) : 3x – y =0}
∴ R= {1,3), (2, 6), (3, 9), (4, 12)}
Here (1, 1) ∉ R, (2, 2) ∉ R
R is not reflexive
Here (1, 3) ∈ R but (3, 1) ∉ R
∴ R is not symmetric
Here (1,3) and (3, 9)6 ∈ R but (1,9) ∉ R
∴ R is not transitive
∴ R is not an equivalence relation.

Question 26.
Solve for x, if tan-1 2x + tan-13x = $$\frac{\pi}{4}$$, x > 0
Given tan-1(2x) + tan-1(3x) = $$\frac{\pi}{4}$$
tan-1 $$\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)$$ = $$\frac{\pi}{4}$$
$$\frac{5 x}{1-6 x^{2}}$$ = tan $$\frac{\pi}{4}$$
⇒ $$\frac{5 x}{1-6 x^{2}}$$ = 1 ⇒ 5x = 1 – 6x2
∴ 6x2 + 5x – 1 = 0
6x2 + 6x – x – 1 =0
6x(x+1) -1(x+1) = 0 or (x+1) (6x – 1)= 0
⇒ x = -1 or x = $$\frac { 1 }{ 6 }$$ Question 27.
By using elementary transformations, find the inverse of the matrix   Question 28.
If x = a(θ + sinθ), y = a( 1 – cosθ) then show that $$\frac{d y}{d x}$$ = tan $$\left(\frac{\theta}{2}\right)$$.
we have
$$\frac{d x}{d \theta}$$ = a(1 + cosθ), $$\frac{d y}{d \theta}$$ = a(sin θ) Question 29.
Verify mean value theorem for the function f(x) = x2 – 4x – 3 in the interval [1, 4].
f(x) = x2 – 4x – 3 interval [a, b] = [1,4]
Here f is a polynomial function hence continuous in [1, 4] and differentiable in (1,4)
f(x) = x2 – 4x – 3 (diff wrt x)
∴ f1(x) = 2x – 4 1
Also f(1) = 12 – 4 – 3 = -6
f(4) = 42 – 4(4) – 3 = -3
∴ From MVT ∃ c ∈ (a,b) such that
f1 (c) = $$\frac{f(b)-f(a)}{b-a}=\frac{-3-(-6)}{4-1}$$
f1 (c) = $$\frac { 3 }{ 3 }$$ = 1
f1 (c) = 2c – 4
∴ 2c – 4 = 1
or 2c = 5
c = $$\frac { 5 }{ 2 }$$ ∈ (1, 4) Question 30.
Find the point, at which the tangent to the curve y = $$\sqrt{4 x-3}-1$$ has its slope $$\frac { 2 }{ 3 }$$
y = $$\sqrt{4 x-3}-1$$ (diff wrt x)
$$\frac{d y}{d x}=\frac{1}{2 \sqrt{4 x-3}} \cdot 4$$
Given slope of tangent = $$\frac{d y}{d x}=\frac{2}{3}$$
∴ $$\frac{2}{3} =\frac{1}{2 \sqrt{4 x-3}} .4$$
$$\frac{4}{3} =\frac{4}{\sqrt{4 x-3}} \Rightarrow \sqrt{4 x-3}=3$$
$$\sqrt{4 x-3} = 3$$ (squaring)
4x – 3 = 9
4x = 12
0r x = 3
∴ y = $$\sqrt{4(3)-3-1}$$
= $$\sqrt{9}$$ – 1 = 3 – 1 = 2
∴ point = (3, 2)

Question 31.
Find $$\int \frac{x}{(x+1)(x+2)} d x$$.  Question 32.
Evaluate $$\int_{0}^{2}\left(x^{2}+1\right) d x$$. as a limit of a sum.
$$\int_{0}^{2}\left(x^{2}+1\right) d x$$ Here a = 0 ; b = 2
f (x) = x2 + 1   Question 33.
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x – axis in the first quadrant.  Question 34.
If y = ae3x + be-2x represents family of curves, where a and b are arbitrary constants. Form the differential equation.
Given, y = ae3x + be-2x …. (1)
y’ = (ae3x)3 – (be-2x) 2
y’ = 3a e3x – 2be-2x …. (2)
From (1) and (2)
y = ae3x + be-2x × 3
y’ = 3a e3x – 2b e-2x × 1 Diff
⇒ 3y’ – y” = 5be-2x
3y’ – y”= -10be-2x …. (4)
From (3) & (4)
3y’ – y” = (-2) (5be-2x)
3y’ – y” = -2[3y – y’]
3y’- y” = -6y + 2y’
y” – y’ – 6y= 0. Question 35.
Show that the position vector of the point P, which divides the line joining the points A and B having position vectors $$\overrightarrow{\mathrm{a}}$$ and $$\overrightarrow{\mathrm{b}}$$ internally in the ratio
$$\mathbf{m}: \mathbf{n} \text { is } \frac{\mathbf{m} \overrightarrow{\mathbf{b}}+\mathbf{n} \overrightarrow{\mathbf{a}}}{\mathbf{m}+\mathbf{n}}$$
Here P divides AB internally in the ratio m : n ∴ p divides $$\overrightarrow{\mathrm{AB}}$$ such that
$$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{m}}{\mathrm{n}}$$ or n$$\overrightarrow{\mathrm{AP}}$$ = m$$\overrightarrow{\mathrm{PB}}$$
Where m and n are positive scalars
Here $$\overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$$ are position vectors of $$\overrightarrow{\mathrm{OP}}$$
$$\overrightarrow{\mathrm{OA}}$$ an $$\overrightarrow{\mathrm{OB}}$$ respectively
∴ From the figure we have, Question 36.
Prove that $$[\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}]$$ = 2$$[\vec{a}, \vec{b}, \vec{c}]$$
LHS  Question 37.
Find the shortest distance between the lines $$\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$$ $$\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$$
2nd PUC Maths Previous Year Question Paper June 2019 27

Question 38.
Bag I contains 3 red and 4 black balls. White Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find
Let A be the event of choosing the bag I
B be the event of choosing the bag II
and C be the event of drawing a red ball
∴ P(A) = P(B) = $$\frac { 1 }{ 2 }$$
Also p(C/A) = $$\frac { 3 }{ 7 }$$ and P(C/B) = $$\frac { 5 }{ 11 }$$
∴ P(C)=P(A).P(C/A) +P(B).P(C/B)
$$=\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}=\frac{3}{14}+\frac{5}{22}=\frac{33+35}{154}$$
P(C) = $$\frac { 68 }{ 154 }$$
By Baye’s Theorem, P(B/C) = $$\frac { 35 }{ 68 }$$ Part-D

Answer any SIX questions: (6 × 5 = 30)

Question 39.
Prove that the function f = R → R defined by f(x) = 4x + 3, is invertible and find the inverse of f.
Let f(x1) = f(x2)= 4x1 + 3 ⇒ 4x2 + 3
⇒ 4x1 = 4x2 ⇒ x1 = x1
f is onto.
Let y ∈ R∃ x ∈ R such that f(x) = y
4x + 3 = y $$x=\frac{y-3}{4} \in R$$
∴ f is onto ∴ f is bijection
∴ f is invertible
Define a function g : R → R by
g(y) = $$\frac{y-3}{4}$$
(fog) (y) = f(g(y)) = f$$\frac{y-3}{4}$$
= 4$$\frac{y-3}{4}$$+3 = y – 3 + 3 = y ∴ (fog) (y) = y fog = I
(gof) (x) = g(f(x)) = g (4x + 3)
$$=\frac{4 x+3-3}{4}=\frac{4 x}{4}=x$$
∴ (gof) (x) = x ∴gof = I
∴ f-1(y) = g(y) = $$\frac{y-3}{4}$$ ∀ y ∈ R

Question 40.
If  Calculate AC, BC and (A + B)C.  Question 41.
Solve the following system of linear equations by matrix method:
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
The system of equations can be written in the form A X = B where |A| = 3 (2 – 3) + 2(4 + 4) + 3 (-6 -4) = – 17
Hence, A is non singular and so its inverse exists. Now
A11 = -1, A12 = -8 A13 = -10
A21 = -5, A22 = -6 A23 = +1
A31 = -1, A32 = +9 A33 = +7
A-1 = $$\frac{1}{|\mathrm{A}|}$$ adj A substituting in (1)  Question 42.
If y = 3 cos (logex) + 4 sin (logex) show that x2y2 + xy1 + y = 0.
y = 3 cos (logex) + 4sin (logex)
Diff wrt x
y1 = 3(-sin(logx).$$\frac { 1 }{ x }$$) + 4 cos (log x). $$\frac { 1 }{ 2 }$$
xy1 = -3 sin (logx) + 4 cos(logx)
diff wrt x
xy2 y1. 1 = -3cos(logx).$$\frac { 1 }{ x }$$ + 4(-sin(logx)).$$\frac { 1 }{ x }$$
xy2 + y1 = $$\frac { 1 }{ 2 }$$-(3cos(logx) + 4sin(logx))
x2y2 + xy1 = -(3cos(logx) + 4sin(logx))
x2y2 + xy1 = -y
∴ x2y2+ xyj1+ y = 0

Question 43.
Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one – sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Volume, V = $$\frac{\pi r^{2} h}{3}$$
given; h = $$\frac { 1 }{ 6 }$$ r (or) r = 6h
∴ v = $$\frac{\pi}{3}$$ × (6h)2 ×h = 12πh3
v = 12πh3
diff w.r.t ‘t’ Question 44.
Find the integral of $$\sqrt{x^{2}+a^{2}}$$ with respect to x, and hence find $$\int \sqrt{x^{2}+2 x+5} d$$  Question 45.
Using integration find the area of the region in the first quadrant enclosed by the x – axis, the line y = x, ad circle x2 + y2 = 32 x2 + y2 = 32 and centre (0,0)
∴ radius = r = $$\sqrt{32}$$ = $$4 \sqrt{2}$$
Equation are y = x ….(1) and x2 + y2 = 32 …..(2)
∴x2 + x2 = 32 (usinig (1) and (2))
∴ 2x2 = 32 ⇒ x2 = 16 or x = ±
since region is in first quadrant
x = 4 ⇒ y = 4
∴ Required Area = Area OBM + Area MBA Question 46.
Find the general solution of the differential equation (x+ y)$$\frac{d y}{d x}$$ = 1
(x + y)$$\frac{d y}{d x}$$ = l dx
(x + y)dy = dx or $$\frac{d y}{d x}$$ = x + y
∴ $$\frac{d y}{d x}$$ -x = y which i s in the form
$$\frac{d y}{d x}$$ + Px =Q dy
Here P = -1; Q = y
∴ Integrating Factor = IF = $$e^{\int P d y}$$
$$=e^{\int(-1) d y}=e^{-y}$$
∴ General solution is
x(IF) = $$\int y \cdot(I F) d y+C$$ (Using integration by parts)
= y(-e-y) – $$\int-\left(e^{-y}\right)$$.1.dy + C
= -ye-y +$$\int e^{-y} d y$$ + C ÷ e-y
xe-y = -ye-y – e-y + C
⇒ x = -y-l+C.ey
Or x + y + 1 = C.ey Question 47.
Derive the equation of a plane in normal form both in the vector and Cartesian form. Consider a plane whose perpendicular distance from the origin is d. where d ≠ 0.  This is the Required Equation of a plane m normal form.
To find P and Q :
x2 = 4y and x =4y – 2
=>(4y – 2)2 = 4y ⇒ 16y2 + 4 – 16 y=4 y
⇒ 16y2 – 20y +4 = 0
⇒ (y – 1) (16y – 4) = 0
⇒ y = 1 (or) y = $$\frac { 1 }{ 4 }$$
if y = 1 ; x = 2 ⇒ 2 (2,1)
and y = $$\frac { 1 }{ 4 }$$ ; x = -1 ⇒ (-1, $$\frac { 1 }{ 4 }$$)
Thus the points of intersection are
(2,1) I(-1 .$$\frac { 1 }{ 4 }$$)
∴ Area of the Region O B Q C = Area of BCQ + Area of QCO  Question 48.
A die is thrown 6 times. If ‘getting an odd number’ is success, what is the probability of
a) 5 successes
b) at least 5 successes
c) at most 5 Successes
Let P = probability of getting an odd number = $$\frac { 3 }{ 6 }$$ = $$\frac { 1 }{ 2 }$$
q = 1 – p = 1$$\frac { -1 }{ 2 }$$ = $$\frac { 1 }{ 2 }$$
& n =6
we have P(x) = ncx Px qn-x    Part-E

Answer any ONE question: (1 × 10 = 10)

Question 49.
a) Prove that $$\int_{0}^{b} f(x) \cdot d x=\int_{a}^{b} f(a+b-x) d x$$ and hence. Evaluate $$\int_{\pi / 3}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$$ (June – 2014,Q.No.49a)
Solution : let t = a + b – x then dt = -dx
when x =a and x = b
t = b t = a    f(x) is continuous, find the value of k.   Question 50.
(a) A factory manufactures two types of screws, A and B. Each type of Screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at most 4 hours on any day. The manufacture can sell a package of screws A at a profit of Rs.7 and seres B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit. ∴ Maximise Z = 7x + 10y subject to constraints
4x + 6y ≤ 240
or 2x + 3y ≤ 120 …… (1)
and 6x + 3y ≤ 240
or 2x + y ≤ 80 …… (2)
and x, y ≥ 0
Converting inequalities to equalities
2x + 3y= 120 2x + y = 80 Plot the graph  Maximum profit is Rs.410 which occurs at point B (30, 20)
∴Manufacturer should manufacture 30 packages of screws A and 20 packages of screws B.    